Let's say we have a circle,
and then we have a
diameter of the circle.
Let me draw my best diameter.
That's pretty good.
This right here is the diameter
of the circle or it's a
diameter of the circle.
That's a diameter.
Let's say I have a triangle
where the diameter is one side
of the triangle, and the angle
opposite that side, it's
vertex, sits some place
on the circumference.
So, let's say, the angle or the
angle opposite of this diameter
sits on that circumference.
So the triangle
looks like this.
The triangle looks like that.
What I'm going to show you
in this video is that
this triangle is going
to be a right triangle.
The 90 degree side is going
to be the side that is
opposite this diameter.
I don't want to label it
just yet because that would
ruin the fun of the proof.
Now let's see what we
can do to show this.
Well, we have in our tool kit
the notion of an inscribed
angle, it's relation to
a central angle that
subtends the same arc.
So let's look at that.
So let's say that this is an
inscribed angle right here.
Let's call this theta.
Now let's say that
that's the center of
my circle right there.
Then this angle right here
would be a central angle.
Let me draw another triangle
right here, another
line right there.
This is a central
angle right here.
This is a radius.
This is the same radius
-- actually this
distance is the same.
But we've learned several
videos ago that look, this
angle, this inscribed angle,
it subtends this arc up here.
The central angle that subtends
that same arc is going
to be twice this angle.
We proved that
several videos ago.
So this is going to be 2theta.
It's the central angle
subtending the same arc.
Now, this triangle right here,
this one right here, this
is an isosceles triangle.
I could rotate it and
draw it like this.
If I flipped it over it would
look like that, that, and then
the green side would
be down like that.
And both of these sides
are of length r.
This top angle is 2theta.
So all I did is I took it
and I rotated it around to
draw it for you this way.
This side is that
side right there.
Since its two sides are equal,
this is isosceles, so these to
base angles must be the same.
That and that must be the same,
or if I were to draw it up
here, that and that must be
the exact same base angle.
Now let me see, I already
used theta, maybe I'll
use x for these angles.
So this has to be x,
and that has to be x.
So what is x going
to be equal to?
Well, x plus x plus 2theta
have to equal 180 degrees.
They're all in the
same triangle.
So let me write that down.
We get x plus x plus 2theta,
all have to be equal to 180
degrees, or we get 2x plus
2theta is equal to 180 degrees,
or we get 2x is equal
to 180 minus 2theta.
Divide both sides by 2, you get
x is equal to 90 minus theta.
So x is equal to
90 minus theta.
Now let's see what else
we could do with this.
Well we could look at this
triangle right here.
This triangle, this side over
here also has this distance
right here is also a
radius of the circle.
This distance over here we've
already labeled it, is
a radius of a circle.
So once again, this is also
an isosceles triangle.
These two sides are equal,
so these two base angles
have to be equal.
So if this is theta,
this is also going to
be equal to theta.
And actually, we use that
information, we use to actually
show that first result about
inscribed angles and the
relation between them and
central angles subtending
the same arc.
So if this is theta, that's
theta because this is
an isosceles triangle.
So what is this whole
angle over here?
Well it's going to be theta
plus 90 minus theta.
That angle right there's
going to be theta
plus 90 minus theta.
Well, the thetas cancel out.
So no matter what, as long as
one side of my triangle is the
diameter, and then the angle or
the vertex of the angle
opposite sits opposite of
that side, sits on the
circumference, then this angle
right here is going to be a
right angle, and this is going
to be a right triangle.
So if I just were to draw
something random like this --
if I were to just take a point
right there, like that, and
draw it just like that,
this is a right angle.
If I were to draw something
like that and go out like
that, this is a right angle.
For any of these I could
do this exact same proof.
And in fact, the way I drew it
right here, I kept it very
general so it would apply
to any of these triangles.